Hi,Now a power source from 2.7 to 5 V can be used. If you put some linear voltage regulator from 3.3V or 5V - then the input voltage can be till 35V. With some minimal changes the power supply can be reduced to 2.5V. With more drastic schematic changes the power supply can be reduced to 1.8V. This requires change of the used chips and also in the control approach.

Hi,Try to use rail to rail single supply op amp in inverting configuration between the voltage input chain and the ADC input. The non inverting input connect to a stable filtered voltage reference close to vdd/2, and set the gain such that if your input volatile is -55V the opamp output trends to vdd, and if your input is voltage is +55Vm the output goes close to gnd. You have to adjust the reference voltage a little to make the output swing symmetrical. I would advice to use spice simulator (like LtSpice - free download from the site of Analog devices. There you can find a lot of opamp models, which can be used for this purpose)

Hi,The right one could work. I would remove R2 if possible.Practically this measurement is not fully identical with the 4 probe measurement. There the main idea is to pass some known current and to measure the voltage drop, here the current is defined by your resistors and the applied voltage.If you remove the R2 than you can measure precisely the voltage asVin = (R6+R7)/R7*Vmeas, where Vin is the battery voltage (input voltage), Vmeas is the sampled voltage.The current sourced by the battery can be estimated as Imeas+Vin/(R6+R7),where Imeas is the measured current and Vin is the calculated voltage(see above).If R2 is your load and you can not omit it, then the formulas will change.R2 must be known and have to be inserted in the calculations.What is the tool that you use for drawing the...

Hi,The right one could work. I would remove R2 if possible.Practically this measurement is not fully identical with the 4 probe measurement. There the main idea is to pass some known current and to measure the voltage drop, here the current is defined by your resistors and the applied voltage.If you remove the R2 than you can measure precisely the voltage asVin = (R6+R7)/R7*Vmeas, where Vin is the battery voltage (input voltage), Vmeas is the sampled voltage.The current sourced by the battery can be estimated as Imeas+Vin/(R6+R7),where Imeas is the measured current and Vin is the calculated voltage(see above).If R2 is your load and you can not omit it, then the formulas will change.R2 must be known and have to be inserted in the calculations.What is the tool that you use for drawing the circuit?Do you use some spice simulator?I would suggest you to simulate the circuit before building it.In this way you can observe all voltage and currents and to see their behavior.here is a nice free spice simulator, which you could use , in the case you do not have one : http://www.linear.com/solutions/ltspice

HI,as I see on your circuit, the resistor R1 I connected from both sides to Arduino GND - that means it is shorted. No voltage drop exist over it.That is the reason your voltage is always zero.Re-arrange the circuit - I would suggest to put R1 from the left side of R3.You will have some small voltage measurement error, but you can cancel it by the software. Also the current should affected by the additional voltage divider R6,R7...but this can be also corrected.

Hi,In the state it is now - no.hardware or software changes must be implemented.Software - you could make FFT and find the signal amplitudeHardware - the voltage must be rectified, filtered and measured as DC - further calculations must be done to receive the RMS value.

Hi,I did not produce current. It is flowing by itself. Imagine a wire through which a current is flowing. We cut the wire and put the ampermeeter between two cut parts. That means - we short again both pieces but through 1Ohm resistor. (we suppose it does not effect the current value). The current flows again. But it creates a voltage drop over the 1Ohm resistors. That is what we measure.

The opamp is supplied by the common supply (5V). Its ground is the common ground. The amperemeter is used in series as usual amperemeter. Please read step to for more precise explanation.

Hi,The input for the amperemeter is current, but there is a current to voltage conversion inside the DMM - The current passing through 1 Ohm resistor creates voltage drop. And practically what the DMM does is measuring this voltage, but not the actual current. You can see the schematics of the amperemeter at step 5. Because the voltage is measured (which is product of the current passing through 1 Ohm resistor), there is not a big difference between the voltmeter and the amperemeter in the code.Only the calculations of the results, which shall be shown on the screen are different. The units also.The procedure performing current measurement is A_500() - you can scroll the code and find it easy.

Hi,Practically all amperemeters work in similar way - the current which must be measured is passed through small resistor. It creates a voltage drop over the resistor, and this voltage drop is measured by the embedded in the multimeter voltmeter. Further a calculation of the current is done : known are the resistor value and the voltage drop over the resistor and according the Ohm's law I=V/R.In my case I use the same approach - the measured current is passed through 1 Ohm resistor. After that I use amplifier by 10 - I multiply the voltage drop over the resistor 10 times. For that purpose an operational amplifier in noninverting configuration is used (the input resistance of this configuration is practically infinity).So a current of 500 mA flowing through 1 Ohm resistor creates 0.5V v...

Hi,Practically all amperemeters work in similar way - the current which must be measured is passed through small resistor. It creates a voltage drop over the resistor, and this voltage drop is measured by the embedded in the multimeter voltmeter. Further a calculation of the current is done : known are the resistor value and the voltage drop over the resistor and according the Ohm's law I=V/R.In my case I use the same approach - the measured current is passed through 1 Ohm resistor. After that I use amplifier by 10 - I multiply the voltage drop over the resistor 10 times. For that purpose an operational amplifier in noninverting configuration is used (the input resistance of this configuration is practically infinity).So a current of 500 mA flowing through 1 Ohm resistor creates 0.5V voltage drop x10 = 5V(as big as the supply voltage of the arduino and as its reference voltage used .for the analog to digital conversion.). Amplifying the voltage drop over the resistors makes the resolution of the measured voltage 10 times better.As you know the arduino main chip has embedded 10 bit ADC.The amplified voltage is measured by it - using the AnalogRead() command. I measure it 16 times. Then I sum all these results and divide the sum by 16. In this way I receive some average value, what improves the accuracy of the measurement - removes the noise in some degree and improves the resolution. This value is still digital word (big from 0 to 1023). And finally I calculate the real current:the digital word I divide by 1024 - this gives the ratio measured voltage divided by reference voltage. After that the division is multiplied by the reference voltage (in our case the ATMEGA328 supply voltage) and we have the needed measured voltage. We divide it by 10 to cancel the amplification, and divide it by 1 Ohm (in my case - it can be also different value - depends on the resistor used) and we have finally the needed current.

The input resistance depends on the potentiometers used.The input resistance of the mixer amplifier is 20k (for audio). As it is, it can work also with line level signals. If the gain is not enough - you can change it for both stages (increasing the feedback resistors), but you have to check what is the recommended gain for the output amplifier. If you want to increase drastically the input resistance, you can change the configuration of the input stage to non-inverting. The output impedance of the mixer is low. Not more than 100 Ohm for 20 Hz signal. It can drive both - headphones and line.

From left side these caps see the vdd/2 voltage.If no signal applied from right side they see the ground potential through the pot.Of course, the potential from left side cna become more positive, but normally the signal source is also AC coupled, which defines the DC potential at the capacitors left side always to 0.

That is true. You have to create virtual ground. In this case we use single supply of 5 V. The best voltage for the virtual ground should be 2.5V.That means - you connect two identical 5 to 10 K resistors in series between the 5V and the ground. At the middle point you put capacitor (1 uF) and connect this node to the pins 3 and 5 of the opamp as virtual ground. That will force the AC signal to dance around this voltage :-).Seems that mcp6022 is a nice alternative for this purpose.

Hi MartinI34,You can remove the headphone amp.You have to create additional reference voltage for the first amplifier - voltage divider - supply/2 and to keep the filtering cap (C11).C9 and C10 prevent flowing of the DC current through the load (the headphones) - they pass only the AC (the signal). If you omit the headphone amplifier, you have to use also such AC coupling capacitors between the mixer and your device - you can take the signal after the capacitors C7,C8.

Practically the POF does not contain any metal wire to attract the lightnings and it does not conduct electrical current. In practically impossible case, when the POF cable is wet it could be conductive, but in this case everything around would be...I do not know how protected is the cable against the sunlight - may if exposed to direct sun rays, its external plastic shield could become breakable - in this case may you have to use some plastic tube. But I think that it would be not needed - the cable by itself should enough resistant.

The varistors are for high voltages. In all cases the simplest solution is to use some 5.6V zenner diode.

Sorry,I did a mistake.None of the listed chips will work as input voltage protection for the DMM if connected as shown in their datasheets.Problems:1) They must be supplied by the common DMM supply - not by the input voltage, because it could drop even to 0V and the current is limired.2) If only NMOS or PMOS switch used - it will cut some part of the input voltageswhen PMOS used - it will pass only the voltage (rough said) >2.5V, if NMOS only used it will pass voltages <2.5V.So called transmission gate must be used ( parallel connection of NMOS and PMOS) switches controlled by the signals with opposite polarity.That means - at the output of (let's say) NCP346 additional inverter must be connected and it has to control an NMOS switch connected in parallel with the PMOS switch cont...

Sorry,I did a mistake.None of the listed chips will work as input voltage protection for the DMM if connected as shown in their datasheets.Problems:1) They must be supplied by the common DMM supply - not by the input voltage, because it could drop even to 0V and the current is limired.2) If only NMOS or PMOS switch used - it will cut some part of the input voltageswhen PMOS used - it will pass only the voltage (rough said) >2.5V, if NMOS only used it will pass voltages <2.5V.So called transmission gate must be used ( parallel connection of NMOS and PMOS) switches controlled by the signals with opposite polarity.That means - at the output of (let's say) NCP346 additional inverter must be connected and it has to control an NMOS switch connected in parallel with the PMOS switch controlled directly by the "OUT" terminal of the chip.So.... the circuit becomes more and more complicated :-)

Hi,I took a look on the datasheets.The serial diode is optional.You can use NCP346 without it and without problems.

May be NCP346 is not the most suitable circuit for a voltmeter.It has to have only low Ron switch. Check the others.The serial diode would effect the correctness of the measurement because of the voltage drop on it.Normally each multimeter should have some protection on the input.The protection which I use is the simplest and may be most used.As common the DMM devices use dedicated chips (ASIC) and I suppose the protection is embedded directly inside them - in all cases it has to based on some voltage clamping circuit, or circuit similar to mentioned before chips.

Hi ArezkiLAs I remember I bought 50 pcs 5.6V zenner diodes and I have chosen one of them having the smallest reverse current at 5V.There are also another possibilities to protect the input against HV.There exist special overvoltage protection circuits containing MOS switch with very low Ron, which observe the input voltage and if it becomes higher then desired interrupt the connection. All depends on that how much you want to complicate the device :-)Examples of such circuits: NCP346, STBP112,NUS3046MN...and a lot of others.You can use 5% resistors, The introduced error you have to correct further by the software trimming.RegardsMilen

Hi,you can not feed current inside the analog pin. You can apply only voltage - not more than the supply voltage+0.5V.The maximum frequency is determined by the max sample rate:76.9kSPS of the ATMEGA328 ADC and the Nyquist requirement ----> around 38,5 kHz.You can put a fuse from the type that you like, but its max current must be proper chosen

Hi,as designed it measures 500 mA max current - that should be the fuse rating. If you want to measure higher current ´then you have to change also the resistor. You could add also input for higher currents with different resistor and can use also different arduino analog input.

Hi Witgood,I never went so deep inside the hardware, except for the power supply part.I could try to find out which MPU/MCU is used. Also if any operation system is installed. This could be Openwrt. Using an USB to TTL serial converter ((based on FT232, PL2303, CH340G or etc.) you could try to communicate with the board - you have to search for (TX,RX) labels on the board and try to connect there.As terminal you could use Putty. May be you have to try different communication speeds. If you find some linux installation - you could try to update the firmware...You could also try to run some memory tests...It could be an interesting play.Please, if you find some additional info - share it here to be available for all interested.RegardsMilen

Hi,You simply connect it to the serial port of the PC. (IT MUST BE REAL RS232 PORT). It doesn't require any else except the software.It uses Ponyprog : http://www.lancos.com/prog.htmlIn the list of the supported microcontrollers you can find also AT90S1200.

Hi FrancoIn my circuit you could remove R4,D2,C2 and instead them to connect a digital signal which would modulate the laser light. The problem here is that the used parts does not have enough bandwidth to guarantee the high speed modulation. But in all cases the main principle for laser modulators remains similar.Here you can read how the modulation speed is calculated:http://www.loreti.it/Download/PDF/laser/laser_printer.pdfThere are a dedicated high speed chips used for the modulating of the laser light:http://www.ti.com/product/onet4201ldhttp://cds.linear.com/docs/en/datasheet/5100fs.pdfIn the datasheets you can see typical connection circuits of them.One possible technology of the pattering could be the use of DLP.More info: http://www.ti.com/lsds/ti/analog/dlp/overview.pageRegards...

Hi FrancoIn my circuit you could remove R4,D2,C2 and instead them to connect a digital signal which would modulate the laser light. The problem here is that the used parts does not have enough bandwidth to guarantee the high speed modulation. But in all cases the main principle for laser modulators remains similar.Here you can read how the modulation speed is calculated:http://www.loreti.it/Download/PDF/laser/laser_printer.pdfThere are a dedicated high speed chips used for the modulating of the laser light:http://www.ti.com/product/onet4201ldhttp://cds.linear.com/docs/en/datasheet/5100fs.pdfIn the datasheets you can see typical connection circuits of them.One possible technology of the pattering could be the use of DLP.More info: http://www.ti.com/lsds/ti/analog/dlp/overview.pageRegardsMilen

Hi CharleneD16 ,You can replace the chip with a pair of PMOS transistprs (like BSS84 :http://www.nxp.com/documents/data_sheet/BSS84.pdf) Exactly this is SND, but you can find similar in through-hole package. There should not be any difference in the performance.The transistors should be connected in the following way:sources - supplygates - first to the input of the inverter, second at the its outputdrains - to the trimmer potentiometersRegardsMilen

Hi .Please write the type of both sensors. For the temperature sensor I suppose you have to serch the proper arduino library. The voltage sensor seems to me to have an analog output (check if true). In this case you have simply to use AnalogRead(A0). Check what should be the output voltage if you supply the sensor with 9V battery (as on the picture). If it is higher than 5V - an error will occur - you have to insert proper resistor voltage divider at the output of the voltage sensor and the analog arduino pin.

Hi,Check the connections of the voltage sensor - seems that the sig and supply pins are swapped. Is it not possible to supply this sensor with 5V arduino supply?Do not forget pull-up resistors needed for the I2C communication. I suppose that also the rx TTL pin shall be connected - some return data is send also during the communication.regardsMilen

Hi,The feedback resistors should be these placed between pins 1 - 2 and 6-7.May be they are SMD devices. Try to unsolder them and to measure their value with a ohmmeter. After that you could put new ones with higher value - but not more than 1.5-2 times.The maximum supply allowed is 5.5V. 6 V is the absolute maximum rating - applying this voltage for a long time could damage the amp and it is not guaranteed that it can have the full functionality when supplied with 6V.May be better solution could be 3 batteries 1.5V, or as in my case to add a voltage regulator from the type 7805. Then you could apply voltage till 30V.RegardsMilen

You could try to boost the gain of the op-amp changing the feedback resistors if possible.

I am glad that I could attract another guy in the world of the electronics.You can identify the amplifier chip and find it datasheet. Some of the used chips are simply internally defined as buffers (no gain - only source-load resistance matching) - in this case nothing can be done. But is a standard power operational amplifier is used as amplifier stage, normally its gain is fixed by the feedback resistors. (these which connect the output of the amplifier with the corresponding input). If you increase their value - you also increase the gain. You should not increase them a lot because the amplifier can saturate (clip at the supply rails) or the gain increase can cause stability problems (the amplifier can start to oscillate).

Thanks,I think - you have to decide which type of communication you want to use : SPI or I2C and depending on this to define the used pins and libraries. From the code you can remove then the LCD and serial monitor part.

Hi, the schematic can be found on step 7. If you do not want to use the kit linked here, the schematic with the LM317 regulator can be seen at step 1. All other connections are described in the text. Some of them depend on which regulator shall be used or which digital potentiometer. The design is done in the way that it allows a big flexibility and the schematics can be different in each particular realization.

Hi JXplicits,RC4558 could be used, but it requires min. 10V supply. The LM4880 will not work at 10V. You could use additional regulators, but this will make the design more complicated. If you want to use the PCB without changing it, you have to keep also the package type of the active mixer op-amp. One possible solution could be the OPA2316 in MSOP package. I hope you can find it.RegardsMilen